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Question

Let f(n) equals to the sum of the cubes of three consecutive natural numbers.
f(n) leaves the remainder zero when divided by

A
11
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B
9
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C
99
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D
none of these
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Solution

The correct option is C 9
Given that f(n)=(n1)3+n3+(n+1)3=3n3+6n
Put n=1, to obtain f(1)=3.13+6.1=9
Therefore, f(1) is divisible by 9
Assume that for n=k, f(k)=3k3+6k is divisible by 9
Now, f(k+1)=3(k+1)3+6(k+1)=3k3+6k+9(k2+k+1)=f(k)+9(k2+k+1)
Since, f(k) is divisible by 9
Therefore, f(k+1) is divisible by 9
And from the principle of mathematical induction f(n) is divisible by 9 for all nN
Hence, option B is correct.

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